If tan $A = \dfrac{3}{5}$, the value of sin² A + cos² A is :

1. $\dfrac{9}{25}$

2. 1

3. $\dfrac{9}{16}$

4. $\dfrac{16}{9}$

Given:

tan $A = \dfrac{3}{5}$

i.e. $\dfrac{Perpendicular}{Base} = \dfrac{3}{5}$

∴ If length of BC = 3x unit, length of AB = 5x unit.

In Δ ABC,

⇒ AC² = AB² + BC² (∵ AC is hypotenuse)

⇒ AC² = (5x)² + (3x)²

⇒ AC² = 25x² + 9x²

⇒ AC² = 34x²

⇒ AC = $\sqrt{34\text{x}^2}$

⇒ AC = $\sqrt{34}$ x

sin $A = \dfrac{Perpendicular}{Hypotenuse}$ =

$= \dfrac{BC}{AC} = \dfrac{3x}{\sqrt{34}x} = \dfrac{3}{\sqrt{34}}$

cos $A = \dfrac{Base}{Hypotenuse}$

$= \dfrac{AB}{AC} = \dfrac{5x}{\sqrt{34}x} = \dfrac{5}{\sqrt{34}}$

Now, sin² A + cos² A

$= \Big(\dfrac{3}{\sqrt{34}}\Big)^2 + \Big(\dfrac{5}{\sqrt{34}}\Big)^2\\[1em] = \Big(\dfrac{9}{34}\Big) + \Big(\dfrac{25}{34}\Big)\\[1em] = \Big(\dfrac{9 + 25}{34}\Big)\\[1em] = \Big(\dfrac{34}{34}\Big)\\[1em] = 1$

Hence, option 2 is the correct option.